Can gun recoil really slow a fighter?

[parsifal]

It is Newtons 3rd law; The momentum of the "stuff" going out the muzzle HAS to equal the momentum of gun, mount, plane, ship.

Is it momentum (MV) or Force (MA) that we need to look at? Does a cannon ball weighing 100 lbs and travelling at 50m/s generate the same recoil as a bullet weighing 1lb travelling at 5000m/s. Disregard all other extraneous variables like friction and wind resistance for the moment.... If the velocity of that cannon ball is not changing, does it generate any recoil at all? Is a cannon ball that is accelarating as it leaves the barrel generating more less or the same recoil as a cannon ball that leaves the barrel having a stable, unchanging velocity?

 

[CobberKane]

Is it momentum (MV) or Force (MA) that we need to look at? Does a cannon ball weighing 100 lbs and travelling at 50m/s generate the same recoil as a bullet weighing 1lb travelling at 5000m/s.
Disregard all other extraneous variables like friction and wind resistance for the moment.... If the velocity of that cannon ball is not changing, does it generate any recoil at all? Is a cannon ball that is accelarating as it leaves the barrel generating more less or the same recoil as a cannon ball that leaves the barrel having a stable, unchanging velocity?

Yes, the 100lb and the 1lb cannon ball would give the same recoil.
In regards to the second scenario, the cannon ball accelerates from a stationary position relative to the cannon, to a given speed relative to the cannon. This MUST create recoil (Newton's third law of motion). The cannon ball will accelerate while in the barrel, but it will NOT accellerate after it leaves tha barrel as there is no longer any force compelling it to do so. Hence, muzzle velocity, measured as the projectile leaves the muzzle, is the highest speed reached by a projectile and therefore is the measure we would use in calculating recoil. The only exception I can think of to this is with a dud round, where the projectile might be accelerating negativeley when it leaves the muzzle.

 

[Rick65]

Long time since I did physics but isn't the force (recoil) generated a function of the velocity squared of the projectile as it leaves the end of the barrel?
As such a 1lb projectile exiting at 5000m/sec will generate 100 times the force(recoil) of a 100lb projectile exiting at 50m/sec.
Simplistically
F=m(v squared)/2s
where
F= force (recoil)
m = mass of projectile (bullet plus propellant?)
v = exit velocity of projectile having accelerated from a relative velocity of zero
s = length of barrel

Or should I have stayed retired from physics?

 

[Shortround6]

The momentum is not the force (kinetic energy) the projectile has.

And as Newtons 3rd law states BOTH the 'bullet' and the 'gun' are projectiles. the gun just moves a LOT slower

This can be seen or tested by firing different loads out of the same gun. A fast, light bullet can have more "energy" than a slow heavy bullet but the gun will recoil less with the light bullet. This was remarked on quite a bit with the change over from 10-14.7 mm military black powder rifles to the 6-8mm smokeless powder rifles.

as to the recoil we are also back to the momentum and energy question and this is part of the "felt" recoil problem. A 6lb rifle and an 8lb rifle firing the same cartridge will recoil at different speeds. the energy the rifle hits the shoulder with is the result of the mass of the rifle times the velocity of the rifle squared.

BTW "s = length of barrel" has absolutely nothing to do with recoil or even velocity of the projectile in this question. Longer barrels allow for more velocity but that is getting into the subject of internal ballistics (what happens inside the barrel.

You do have to figure the bullet and propellant seperately however as they have vastly different "exit' velocities.

 

[Rick65]

Newton's 2nd F(recoil)= mass x acceleration.
If we assume that a projectile accelerates uniformly until it reaches muzzle velocity as it leaves the barrel (is this correct or do projectile reach maximum velocity sometime before the end of the barrel?), the length of the barrel is significant as it allows us to calculate the acceleration of the projectile as it changes from stationary to moving at muzzle velocity over the length of the barrel.
v squared = u squared +2as
Where
v=end velocity (muzzle?)
u= initial velocity (=0 in this case)
a= acceleration
s = distance traveled (barrel length?)

 

[Shortround6]

But we don't need to calculate the acceleration of the projectile in order to figure out the recoil.


ALL we need is the mass of the "stuff" leaving the barrel and the velocity/s of the different "stuff/s" (projectile and powder gases). It doesn't make any difference if the projectile took 700mm of barrel or 900mm of barrel travel to reach 700 meters a second.

Your formula may be correct but it is solving the wrong question.

 

[syscom3]

The formula to use is the kinetic energy (in joules) of the airplane of "x" weight at "Y" speed. And in the other direction, the force created by the kinetic energy (in joules) of the projectiles of "a" weight going at "b" velocity at the muzzle.

And if the ratio between the two is high, you can bet that the speed of the aircraft that is lost , would be quite low.

 

[Rick65]

Thanks Shortround - much better
Ke=m(v squared)/2
Further proof that a little knowledge (on my part) is dangerous.

 

[OldSkeptic]

You can calculate in different way ... and still pretty much get the same answer. Either a delta KE calc (same as space craft engine calcs) or as an impulse one or as a momentum transfer.

More of an issue in the older planes as they were so much lighter, unless you have really big guns like a A-10.
People forget just how heavy modern planes are. For example a fully loaded F-15C is similar to fully loaded Lancaster Bomber.